Problem: $h(x) = -5x$ $f(n) = n^{3}-7n^{2}-n+1-4(h(n))$ $ f(h(0)) = {?} $
Explanation: First, let's solve for the value of the inner function, $h(0)$ . Then we'll know what to plug into the outer function. $h(0) = (-5)(0)$ $h(0) = 0$ Now we know that $h(0) = 0$ . Let's solve for $f(h(0))$ , which is $f(0)$ $f(0) = 0^{3}-7(0^{2})-0+1-4(h(0))$ To solve for the value of $f$ , we need to solve for the value of $h(0)$ $h(0) = (-5)(0)$ $h(0) = 0$ That means $f(0) = 0^{3}-7(0^{2})-0+1+(-4)(0)$ $f(0) = 1$